Let `A=[{:(a,b,c),(p,q,r),(1,1,1):}]and B=A^(2)`
If `(a-b)^(2) +(p-q)^(2) =25, (b-c) ^(2)+ (q-r)^(2)= 36 and (c-a)^(2) +(r-p)^(2)=49,` then det B is

Options

A. 192
B. 864
C. 2456
D. `25xx36xx47`

Answer

To solve the problem step by step, we will first analyze the given information and then calculate the required determinant.

### Step 1: Understand the Given Information
We have a matrix \( A \) defined as:
\[ A = \begin{pmatrix} a & b & c \\ p & q & r \\ 1 & 1 & 1 \end{pmatrix} \]
We also know that \( B = A^2 \).

We are given three equations:
1. \((a-b)^2 + (p-q)^2 = 25\)
2. \((b-c)^2 + (q-r)^2 = 36\)
3. \((c-a)^2 + (r-p)^2 = 49\)

### Step 2: Calculate the Lengths of the Sides of the Triangle
From the equations, we can interpret them as the lengths of the sides of a triangle formed by the points \((a, p)\), \((b, q)\), and \((c, r)\).

- **Length \( a \)**:
\[ a = \sqrt{(a-b)^2 + (p-q)^2} \]
From the first equation, we have:
\[ a = \sqrt{25} = 5 \]

- **Length \( b \)**:
\[ b = \sqrt{(b-c)^2 + (q-r)^2} \]
From the second equation, we have:
\[ b = \sqrt{36} = 6 \]

- **Length \( c \)**:
\[ c = \sqrt{(c-a)^2 + (r-p)^2} \]
From the third equation, we have:
\[ c = \sqrt{49} = 7 \]

### Step 3: Calculate the Semi-Perimeter \( s \)
The semi-perimeter \( s \) of the triangle is given by:
\[ s = \frac{a + b + c}{2} = \frac{5 + 6 + 7}{2} = 9 \]

### Step 4: Calculate the Area of the Triangle Using Heron's Formula
Using Heron's formula:
\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values we found:
\[ \text{Area} = \sqrt{9(9-5)(9-6)(9-7)} = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216} = 6\sqrt{6} \]

### Step 5: Relate the Area to the Determinant of Matrix \( A \)
The area of the triangle can also be expressed in terms of the determinant of matrix \( A \):
\[ \text{Area} = \frac{1}{2} \cdot \text{det}(A) \]
Thus, we have:
\[ 6\sqrt{6} = \frac{1}{2} \cdot \text{det}(A) \]
From this, we can find:
\[ \text{det}(A) = 12\sqrt{6} \]

### Step 6: Calculate the Determinant of Matrix \( B \)
Since \( B = A^2 \), we have:
\[ \text{det}(B) = \text{det}(A)^2 \]
Substituting the value of \( \text{det}(A) \):
\[ \text{det}(B) = (12\sqrt{6})^2 = 144 \cdot 6 = 864 \]

### Final Answer
Thus, the value of \( \text{det}(B) \) is:
\[ \text{det}(B) = 864 \]